NCERT Exemplar Solutions Class 11 Physics Kinetic Theory — Very Short, Short and Long Answer Questions Q14–31

NCERT Exemplar Problems, Class 11 Physics — Kinetic Theory, Very Short Answer Questions 14 to 19, Short Answer Questions 20 to 26 and Long Answer Questions 27 to 31. All 18 questions solved. The book numbers them 13.14 to 13.31.

Three ideas carry this page. Temperature measures average translational kinetic energy: $\tfrac12 m\overline{v^2} = \tfrac32 kT$, so $v_{rms} = \sqrt{3kT/m}$ depends on temperature and molecular mass, not on pressure. Energy is shared equally among degrees of freedom: $\tfrac12 kT$ each, which gives the internal energy of any ideal gas or mixture. Molecules travel a mean free path $\lambda = 1/(\sqrt2\,\pi d^2 n)$ between collisions, and the same formula works for anything moving at random, even aeroplanes.

Key insight. Most of these reduce to one of two lines: $\tfrac12 m v_{rms}^2 = \tfrac32 kT$, or $U = \tfrac f2 \mu RT$. Write the right one down first and the numbers follow.

Very Short Answer Questions

Question 14

Calculate the number of atoms in 39.4 g gold. Molar mass of gold is $197\ \text{g mol}^{-1}$.

Solution. The number of moles is $39.4/197 = 0.200$ mol, and each mole holds $N_A = 6.02\times10^{23}$ atoms:

$$N = 0.200 \times 6.02\times10^{23} \approx 1.2\times10^{23}.$$

Question 15

The volume of a given mass of a gas at $27^\circ$C, 1 atm is 100 cc. What will be its volume at $327^\circ$C?

Solution. The pressure is unchanged, so Charles’s law applies with absolute temperatures: $T_1 = 300$ K and $T_2 = 600$ K.

$$V_2 = V_1\frac{T_2}{T_1} = 100 \times \frac{600}{300} = 200\ \text{cc}.$$

The common slip is to use $327/27$ and get a twelvefold increase; gas laws only work in kelvin.

Question 16

The molecules of a given mass of a gas have root mean square speeds of $100\ \text{m s}^{-1}$ at $27^\circ$C and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at $127^\circ$C and 2.0 atmospheric pressure?

Solution. Since $v_{rms} = \sqrt{3kT/m}$, the rms speed depends only on the absolute temperature. Doubling the pressure changes the density, not the speed of the molecules. So

$$v_2 = v_1\sqrt{\frac{T_2}{T_1}} = 100\sqrt{\frac{400}{300}} = \frac{200}{\sqrt3} \approx 115\ \text{m s}^{-1}.$$

Question 17

Two molecules of a gas have speeds of $9\times10^6\ \text{m s}^{-1}$ and $1\times10^6\ \text{m s}^{-1}$, respectively. What is the root mean square speed of these molecules?

Solution. Root mean square means square, average, then take the root:

$$v_{rms} = \sqrt{\frac{(9\times10^6)^2 + (1\times10^6)^2}{2}} = \sqrt{\frac{82}{2}}\times10^6 = \sqrt{41}\times10^6 \approx 6.4\times10^6\ \text{m s}^{-1}.$$

It is larger than the ordinary mean, $5\times10^6\ \text{m s}^{-1}$, because squaring gives the faster molecule more weight.

Question 18

A gas mixture consists of 2.0 moles of oxygen and 4.0 moles of neon at temperature $T$. Neglecting all vibrational modes, calculate the total internal energy of the system. (Oxygen has two rotational modes.)

Solution. Each degree of freedom carries $\tfrac12 RT$ per mole. Oxygen has 3 translational and 2 rotational degrees of freedom, so $\tfrac52 RT$ per mole; neon is monatomic, with only 3 translational, so $\tfrac32 RT$ per mole.

$$U = 2\times\tfrac52 RT + 4\times\tfrac32 RT = 5RT + 6RT = 11RT.$$

Question 19

Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1 Å and 2 Å. The gases may be considered under identical conditions of temperature, pressure and volume.

Solution. The mean free path is $\lambda = \dfrac{1}{\sqrt2\,\pi d^2 n}$. Same $p$ and $T$ means the same number density $n = p/kT$, so $\lambda \propto 1/d^2$:

$$\frac{\lambda_1}{\lambda_2} = \left(\frac{d_2}{d_1}\right)^2 = \left(\frac{2}{1}\right)^2 = 4.$$

The bigger molecule presents four times the collision cross-section, so it travels a quarter as far between collisions.

Short Answer Questions

Question 20

The container shown in Fig. 13.6 has two chambers, separated by a partition, of volumes $V_1 = 2.0$ litre and $V_2 = 3.0$ litre. The chambers contain $\mu_1 = 4.0$ and $\mu_2 = 5.0$ moles of a gas at pressures $p_1 = 1.00$ atm and $p_2 = 2.00$ atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.

The figure shows a box divided into a left chamber ($V_1$, $\mu_1$, $p_1$) and a right chamber ($V_2$, $\mu_2$, $p_2$).

Solution. The container is closed and does no work, so the total kinetic energy of the gas is conserved when the partition is removed. For any amount of ideal gas, $pV = \tfrac23 E$, where $E$ is its total translational kinetic energy. Before mixing,

$$E_{total} = \tfrac32(p_1V_1 + p_2V_2).$$

After mixing, the same energy fills the whole volume $V_1 + V_2$:

$$p(V_1 + V_2) = \tfrac23 E_{total} = p_1V_1 + p_2V_2,$$

$$p = \frac{p_1V_1 + p_2V_2}{V_1 + V_2} = \frac{1.00\times2.0 + 2.00\times3.0}{2.0 + 3.0} = \frac{8.0}{5.0} = 1.60\ \text{atm}.$$

The numbers of moles drop out. That is fortunate, because the data as printed are not self-consistent at a common temperature: $pV = \mu RT$ with these values gives the two chambers different temperatures, about 6 K and 15 K, both absurdly cold for a gas. The energy argument needs only $p$ and $V$, which is why it is the right tool here.

Question 21

A gas mixture consists of molecules of types A, B and C with masses $m_A > m_B > m_C$. Rank the three types of molecules in decreasing order of (a) average K.E., (b) rms speeds.

Solution. (a) In a mixture in equilibrium all the molecules are at the same temperature, and the average translational kinetic energy is $\tfrac32 kT$ whatever the mass. So the three are equal.

(b) Since $\tfrac12 m v_{rms}^2$ is the same for all, $v_{rms} = \sqrt{3kT/m}$ is largest for the lightest molecule. With $m_A > m_B > m_C$, the order of rms speed is the reverse: $C$, then $B$, then $A$.

Question 22

We have 0.5 g of hydrogen gas in a cubic chamber of size 3 cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state? (Hydrogen molecules can be consider as spheres of radius 1 Å.)

Solution. 0.5 g of $\text{H}_2$ is $0.25$ mol, that is $0.25\times6.02\times10^{23} \approx 1.5\times10^{23}$ molecules. Each is a sphere of volume $\tfrac43\pi(10^{-10})^3 \approx 4.2\times10^{-30}\ \text{m}^3$, so the molecules themselves occupy

$$V_{mol} \approx 1.5\times10^{23}\times4.2\times10^{-30} \approx 6.3\times10^{-7}\ \text{m}^3.$$

Taking the gas to start at 1 atm in the $27\ \text{cm}^3 = 2.7\times10^{-5}\ \text{m}^3$ chamber and applying Boyle’s law, the final volume would be

$$V_f = \frac{2.7\times10^{-5}}{100} = 2.7\times10^{-7}\ \text{m}^3.$$

That is less than the volume of the molecules themselves. The ideal gas law assumes the molecules are points with no forces between them, and here they would be packed tighter than touching. So the ideal gas law is not justified in the final state.

The printed key uses $10^{-30}\ \text{m}^3$ per molecule (it drops the factor $\tfrac43\pi$) and gets $2.5\times10^{-7}\ \text{m}^3$, comparable with the final volume. The fuller figure makes the conclusion stronger, not weaker.

Question 23

When air is pumped into a cycle tyre the volume and pressure of the air in the tyre both are increased. What about Boyle’s law in this case?

Solution. Boyle’s law, $pV = \text{constant}$, holds only for a fixed amount of gas at constant temperature. Pumping adds more air molecules to the tyre, so $\mu$ in $pV = \mu RT$ increases, and both $p$ and $V$ can rise together. Boyle’s law is not violated; it simply does not apply, because its condition of a fixed mass is not met.

Question 24

A balloon has 5.0 g mole of helium at $7^\circ$C. Calculate (a) the number of atoms of helium in the balloon, (b) the total internal energy of the system.

Solution. (a) $N = \mu N_A = 5.0\times6.02\times10^{23} \approx 3.0\times10^{24}$ atoms.

(b) Helium is monatomic, so its internal energy is purely translational, $\tfrac32 kT$ per atom, with $T = 280$ K:

$$U = \tfrac32 NkT = \tfrac32\mu RT = \tfrac32\times5.0\times8.31\times280 \approx 1.75\times10^4\ \text{J}.$$

Question 25

Calculate the number of degrees of freedom of molecules of hydrogen in 1 cc of hydrogen gas at NTP.

Solution. One mole of any ideal gas occupies $22\,400$ cc at NTP, so 1 cc holds

$$N = \frac{6.02\times10^{23}}{22\,400} \approx 2.69\times10^{19}\ \text{molecules}.$$

Each diatomic hydrogen molecule has 5 degrees of freedom (3 translational, 2 rotational) at ordinary temperatures, so the total is

$$5N \approx 1.34\times10^{20}.$$

Question 26

An insulated container containing monoatomic gas of molar mass $m$ is moving with a velocity $v_0$. If the container is suddenly stopped, find the change in temperature.

Solution. While the container moves, every molecule carries the ordered velocity $v_0$ on top of its random motion. When it is stopped, that ordered kinetic energy is converted by collisions into random motion, that is, into internal energy; the walls are insulating, so none of it escapes. For $n$ moles the mass is $nm$, so

$$\tfrac12 (nm)v_0^2 = n\cdot\tfrac32 R\,\Delta T \quad\Longrightarrow\quad \Delta T = \frac{m v_0^2}{3R}.$$

The temperature rises.

Long Answer Questions

Question 27

Explain why (a) there is no atmosphere on moon. (b) there is fall in temperature with altitude.

Solution. (a) The Moon’s gravity is weak: its escape speed is $v_{esc} = \sqrt{2GM/R} \approx 2.4\ \text{km s}^{-1}$, against $11.2\ \text{km s}^{-1}$ for the Earth. The Moon is the same distance from the Sun as the Earth, so its sunlit surface is at a few hundred kelvin. At 300 K the rms speed of nitrogen is about $0.52\ \text{km s}^{-1}$ and of hydrogen about $1.9\ \text{km s}^{-1}$. These rms values are below $2.4\ \text{km s}^{-1}$, but the Maxwell distribution has a long tail: a small fraction of molecules always moves faster than escape speed and leaves. The remaining gas then re-establishes the same distribution by collisions, refilling the tail, and more molecules escape. Over the Moon’s billions of years this steady leak has removed almost all of any atmosphere it had. On Earth the escape speed is so much higher that the tail is negligible for everything except the lightest gases.

(b) A gas molecule moving upward gains gravitational potential energy and so loses kinetic energy, and temperature measures kinetic energy. In addition, air that rises reaches lower pressure, expands almost adiabatically and cools. Both effects make the temperature fall with height in the lower atmosphere.

Printed key error. The key’s argument is right but two of its supporting numbers are not. It writes $v_{rms} = \sqrt{3\times1.38\times10^{-23}\times300/(7.3\times10^{-26})} = 1.7\ \text{km s}^{-1}$; that expression actually equals $0.41\ \text{km s}^{-1}$. And it gives the Moon’s escape speed as $4.6\ \text{km s}^{-1}$, whereas $\sqrt{2GM/R}$ with $M = 7.35\times10^{22}$ kg and $R = 1.74\times10^6$ m gives $2.4\ \text{km s}^{-1}$. The conclusion (rms speed below escape speed, escape through the tail) stands.

Question 28

Consider an ideal gas with following distribution of speeds.

Speed (m/s)	% of molecules
200	10
400	20
600	40
800	20
1000	10

(i) Calculate $V_{rms}$ and hence $T$. $(m = 3.0\times10^{-26}\ \text{kg})$ (ii) If all the molecules with speed 1000 m/s escape from the system, calculate new $V_{rms}$ and hence $T$.

Solution. (i) Weight each speed squared by its fraction:

$$\overline{v^2} = 0.1(200)^2 + 0.2(400)^2 + 0.4(600)^2 + 0.2(800)^2 + 0.1(1000)^2 = 4.08\times10^5\ \text{m}^2\text{s}^{-2},$$

so $v_{rms} = \sqrt{4.08\times10^5} \approx 639\ \text{m s}^{-1}$. From $\tfrac12 m\overline{v^2} = \tfrac32 kT$,

$$T = \frac{m\overline{v^2}}{3k} = \frac{3.0\times10^{-26}\times4.08\times10^5}{3\times1.38\times10^{-23}} \approx 296\ \text{K}.$$

(ii) Removing the 10% moving at 1000 m/s leaves 90% of the molecules, with $\sum (\text{fraction})\,v^2 = 3.08\times10^5$:

$$\overline{v^2} = \frac{3.08\times10^5}{0.9} \approx 3.42\times10^5\ \text{m}^2\text{s}^{-2}, \quad v_{rms} \approx 585\ \text{m s}^{-1}, \quad T = \frac{m\overline{v^2}}{3k} \approx 248\ \text{K}.$$

Losing only the fastest molecules lowers the temperature by nearly 50 K. This is cooling by evaporation: the escaping molecules carry away more than their share of energy.

Question 29

Ten small planes are flying at a speed of 150 km/h in total darkness in an air space that is $20\times20\times1.5\ \text{km}^3$ in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius 10 m.

Solution. The planes behave like gas molecules, so use the mean free path. Two safety spheres of radius 10 m touch when their centres are 20 m apart, so the effective diameter is $d = 20$ m. The number density is

$$n = \frac{10}{20\times20\times1.5\ \text{km}^3} = 0.0167\ \text{km}^{-3} = 1.67\times10^{-11}\ \text{m}^{-3}.$$

$$\lambda = \frac{1}{\sqrt2\,\pi d^2 n} = \frac{1}{1.414\times3.14\times400\times1.67\times10^{-11}} \approx 3.4\times10^7\ \text{m} = 3.4\times10^4\ \text{km}.$$

At 150 km/h the time between near collisions is

$$t = \frac{\lambda}{v} = \frac{3.4\times10^4}{150} \approx 225\ \text{h}.$$

Question 30

A box of $1.00\ \text{m}^3$ is filled with nitrogen at 1.50 atm at 300 K. The box has a hole of an area $0.010\ \text{mm}^2$. How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm.

Solution. Follow the book’s simple model. Each molecule has a velocity component along $x$ with $\overline{v_x^2} = \tfrac13\overline{v^2} = kT/m$, so take $v_x = \sqrt{kT/m}$. Half the molecules near the hole move towards it, so in time $\Delta t$ the number leaving through area $a$ is $\tfrac12 n_1 v_x a\,\Delta t$, and the number entering from outside is $\tfrac12 n_2 v_x a\,\Delta t$. The temperature is the same inside and out, so the net outflow rate is

$$\frac{dN}{dt} = \tfrac12 (n_1 – n_2)\sqrt{\frac{kT}{m}}\;a, \qquad n = \frac{p}{kT}.$$

Treating this rate as constant over the small drop in pressure, the number lost from the box in time $\tau$ is $V(n_1 – n_1′)$, so

$$\tau = 2\,\frac{p_1 – p_1′}{p_1 – p_2}\cdot\frac{V}{a}\sqrt{\frac{m}{kT}}.$$

With $p_1 – p_1′ = 0.10$ atm, $p_1 – p_2 = 0.50$ atm, $V = 1.00\ \text{m}^3$, $a = 1.0\times10^{-8}\ \text{m}^2$ and $m = 28\times1.67\times10^{-27} = 4.67\times10^{-26}$ kg:

$$\sqrt{\frac{m}{kT}} = \sqrt{\frac{4.67\times10^{-26}}{1.38\times10^{-23}\times300}} = 3.36\times10^{-3}\ \text{s m}^{-1},$$

$$\tau = 2\times\frac{0.10}{0.50}\times\frac{1.00}{1.0\times10^{-8}}\times3.36\times10^{-3} \approx 1.34\times10^5\ \text{s},$$

about 37 hours.

Printed key note. The key uses this same formula but prints $1.38\times10^5$ s; its own numbers give $1.34\times10^5$ s, so the difference is arithmetic. The model is rough in two places: the pressure difference falls from 0.50 to 0.40 atm during the leak (allowing for that gives about $1.5\times10^5$ s), and a full Maxwell-distribution treatment of effusion gives a somewhat smaller flux. Either way the answer is of order $10^5$ s, a day or two.

Question 31

Consider a rectangular block of wood moving with a velocity $v_0$ in a gas at temperature $T$ and mass density $\rho$. Assume the velocity is along $x$-axis and the area of cross-section of the block perpendicular to $v_0$ is $A$. Show that the drag force on the block is $4\rho A v_0\sqrt{\dfrac{kT}{m}}$, where $m$ is the mass of the gas molecule.

Solution. Let $n$ be the number of molecules per unit volume, and, as in Question 30, give each molecule an $x$-speed $v = \sqrt{kT/m}$ (since $\overline{v_x^2} = kT/m$), half moving in each direction.

Front face. Molecules coming head-on approach the face at relative speed $v + v_0$. In time $\Delta t$ the number that hit area $A$ is $\tfrac12 n(v + v_0)A\,\Delta t$, and each rebounds elastically, reversing its relative velocity, so it delivers momentum $2m(v + v_0)$. The momentum per unit time, the force on the front face, is

$$F_{front} = \tfrac12 n(v+v_0)A \times 2m(v+v_0) = mnA(v + v_0)^2.$$

Back face. The molecules catching up with the back face approach at relative speed $v – v_0$, so by the same argument

$$F_{back} = mnA(v – v_0)^2,$$

pushing the block forward.

Net drag. The front force opposes the motion and the back force helps it, so

$$F = mnA\left[(v+v_0)^2 – (v-v_0)^2\right] = mnA\,(4vv_0) = 4\rho A v_0\sqrt{\frac{kT}{m}},$$

using $\rho = mn$ and $v = \sqrt{kT/m}$. The drag is proportional to $v_0$, which is the familiar viscous-type drag at low speed, and it grows with temperature because the molecules bombard the block harder.

Common mistakes

• Using degrees Celsius in a gas law (Questions 15, 16). Ratios of temperature must be in kelvin.
• Letting pressure change the rms speed in Question 16. $v_{rms} = \sqrt{3kT/m}$ depends on $T$ and $m$ only.
• Averaging speeds instead of squares in Question 17. The rms of $9$ and $1$ is $\sqrt{41}$, not $5$.
• Giving neon rotational energy in Question 18. A single atom has only the 3 translational degrees of freedom.
• Thinking heavier molecules have more kinetic energy in Question 21. At the same temperature the average is $\tfrac32 kT$ for all; heavier ones are simply slower.
• Saying Boyle’s law “fails” for the tyre in Question 23. It does not apply, because the mass of gas changes.
• Taking $d$ as the radius of the safety sphere in Question 29. Two spheres of radius 10 m touch at a centre separation of 20 m.

Practise next

• Problem. At what temperature will the rms speed of hydrogen molecules equal that of oxygen molecules at $47^\circ$C? (Molar masses 2 and 32 g/mol.)

Show answer

$20$ K. Equal rms speeds need equal $T/M$: $T_H/2 = 320/32$, so $T_H = 320\times2/32 = 20$ K, which is $-253^\circ$C.

• Problem. Estimate the mean free path of a gas molecule of diameter $2\ \text{Å}$ at $0^\circ$C and 1 atm.

Show answer

About $2.1\times10^{-7}$ m. $n = p/kT = 101325/(1.38\times10^{-23}\times273) \approx 2.69\times10^{25}\ \text{m}^{-3}$, so $\lambda = 1/(\sqrt2\,\pi d^2 n) = 1/(1.414\times3.14\times4\times10^{-20}\times2.69\times10^{25}) \approx 2.1\times10^{-7}$ m, roughly a thousand molecular diameters.

• Exemplar Kinetic Theory: Multiple Choice Questions — the conceptual questions on wall collisions, gas laws and equipartition.
• NCERT textbook Chapter 12 — Kinetic Theory for the derivation of gas pressure, rms speed, equipartition and mean free path.
• Exemplar Thermodynamics — the previous unit, on internal energy, work and heat.
• Exemplar Oscillations — the next unit.